A simple python list program getting Index error

Question

Sorry for asking this question. This is simple python program

odd = lambda x : bool(x % 2)
numbers = [n for n in range(10)]
print numbers
for i in range(len(numbers)):
    if numbers[i]%2!=0:
        del numbers[i]
    print numbers

I am getting error while running this program Traceback (most recent call last): File "test.py", line 5, in <module> if numbers[i]%2!=0: IndexError: list index out of range

I tried many thing but nothing is working, Can any one help me

it is because you are deleting the elements in loop rather save it in other list — The6thSense, Sep 15 '15 at 07:30
possible duplicate of [Remove items from a list while iterating in Python](http://stackoverflow.com/questions/1207406/remove-items-from-a-list-while-iterating-in-python) — sebastian, Sep 15 '15 at 07:46

The6thSense · Answer 1 · 2015-09-15T07:57:26.823

3

Notes:

If you really NEED TO DELETE THE LIST WHEN LOOPING you could loop the list in reverse and delete the elements it will not stop the iteration since it will delete the element in normal direction (left --> right) and the loop is in opposite direction (right-->left)

Code:

odd = lambda x : bool(x % 2)
numbers = [n for n in range(10)]
print numbers
for i in numbers[::-1]:
    print i
    if i%2!=0:
        del numbers[numbers.index(i)]
print numbers

Output:

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
9
8
7
6
5
4
3
2
1
0
[0, 2, 4, 6, 8]

Or to be even more simple how about a list comprehension and your lambda function

Code1:

[num for num in numbers if not odd(num)]

Output1:

 [0, 2, 4, 6, 8]

edited Sep 15 '15 at 07:57

answered Sep 15 '15 at 07:36

The6thSense

8,103
8
31
65

@khelwood what mistake ? – The6thSense Sep 15 '15 at 07:53
You are using `i` as the items in the list, not the index, so `pop(i)` is not suitable. – khelwood Sep 15 '15 at 07:56

huang · Answer 2 · 2015-09-15T08:06:38.640

2

b = range(10)
a = [i for i in b if not i%2]

print a

edited Sep 15 '15 at 08:06

answered Sep 15 '15 at 08:04

huang

521
3
11

1

Why not just `a = [i for i in b if i%2==0]` – The6thSense Sep 15 '15 at 08:06

score 1 · Accepted Answer · 2015-09-15T07:42:05.737

1

 odd = lambda x : bool(x % 2)
    numbers = [n for n in range(10)]
    print numbers
    for i in range(len(numbers)):
        try:
            if numbers[i]%2!=0:
               del numbers[i]
                print numbers
        except IndexError:
            pass

edited Sep 15 '15 at 07:42

answered Sep 15 '15 at 07:40

A simple python list program getting Index error

3 Answers3